HSC 2015 MX2 Permutations & Combinations Marathon (archive) (2 Viewers)

[Drsoccerball]

Re: 2015 permutation X2 marathon

is everyone elses latex screwing up?

 

[braintic]

Re: 2015 permutation X2 marathon

is everyone elses latex screwing up?

I can see your answer when I "reply with quote".
It's hard to tell without proper formatting, but it seems to be correct.

 

[kawaiipotato]

Re: 2015 permutation X2 marathon

new q:
There are 10 people, and three rooms A, B, C. How many ways can all these ten people accommodate these rooms such that each room can only have a maximum of 4 people?
not sure of the answer but I got: 44100

 

[kawaiipotato]

Re: 2015 permutation X2 marathon

Could you provide the answer for us to check?

Sorry, but it was in a multiple choice that we got today in school. The choices were: 44100, 22050, and two other ones in the ten thousands but I can't remember them

 

[Drsoccerball]

Re: 2015 permutation X2 marathon

new q:
There are 10 people, and three rooms A, B, C. How many ways can all these ten people accommodate these rooms such that each room can only have a maximum of 4 people?
not sure of the answer but I got: 44100

Possible arrangements are : 4,4,2 and 4,3,3 (which obviously can be arranged)

(10C4)(6C4)(2C2)3!/2! + (10C4)(6C4)(2C2)3!/2! = 22050

 

[kawaiipotato]

Re: 2015 permutation X2 marathon

I was conflicted in the MC and couldn't decide whether we divide by 2 or not, so I didn't. I thought you'd divide by 2! only if the way that they are placed don't matter and something like 4,4,2 and 4,4,2 means 8 different people could be in the first two rooms as opposed to the second 4,4,2. Idk if that makes sense

 

[Ekman]

Re: 2015 permutation X2 marathon

I don't think you should divide by 2! because they are distinct rooms labelled A, B, and C

 

[Drsoccerball]

Re: 2015 permutation X2 marathon

I don't think you should divide by 2! because they are distinct rooms labelled A, B, and C

Thats what i thought but remember multiple choice question 10 in or paper both were right

 

[kawaiipotato]

Re: 2015 permutation X2 marathon

I don't think you should divide by 2! because they are distinct rooms labelled A, B, and C

You would only divide by like 3! if a question said: how many ways can 3 pairs of people up to play a game, right? as opposed to how may ways can you pick 3 pairs of people?

 

[Ekman]

Re: 2015 permutation X2 marathon

Thats what i thought but remember multiple choice question 10 in or paper both were right

No the reason why I divided by 2! for that question was because I assumed the tables were identical. However in this question, the rooms are distinct, so you don't divide.

 

[InteGrand]

Re: 2015 permutation X2 marathon

You divide by 2! to make sure you're not overcounting the no. of ways to form the two groups of 4 in 4, 4, 2 and the two groups of 3 in 4, 3, 3.

 

[Drsoccerball]

Re: 2015 permutation X2 marathon

You divide by 2! to make sure you're not overcounting the no. of ways to form the two groups of 4 in 4, 4, 2 and the two groups of 3 in 4, 3, 3.

But as ekman mentioned the rooms are not identical ? Therefore 4 in one room and 4 in the other is different to another 4 in one room and another 4 in the other?

 

[Ekman]

Re: 2015 permutation X2 marathon

You divide by 2! to make sure you're not overcounting the no. of ways to form the two groups of 4 in 4, 4, 2 and the two groups of 3 in 4, 3, 3.

To form the groups of course, but when you put them into distinct rooms, having 4 people in room A and having 4 people in room B, and having them switch are two different arrangements, as 4 that were initially in A are now in B, and vice versa.

 

[braintic]

Re: 2015 permutation X2 marathon

Drsoccerball's solution is correct.

3!/2! = 3, and there most definitely is 3 ways to arrange 4,4,2 or 4,3,3.

Another thought process (but same calculation):

(For the 4,4,2 case):
Pick the 2 first: 10C2
Pick the room for those two: 3
Pick one of the other rooms, and choose the 4 people for it: 8C4

(For the 4,3,3 case):
Pick the 4 first: 10C4
Pick the room for those four: 3
Pick one of the other rooms, and choose the 3 people for it: 6C3

10C2 times 3 times 8C4 + 10C4 times 3 times 6C3 = 22050

 

[kawaiipotato]

Re: 2015 permutation X2 marathon

new q:
There are 'n' cars in a parking lot with 'n' exits. Any car can go into any exit. What is the probability that there is at least one exit which is not accessed by any car?
Again, don't have the solution to this

 

[braintic]

Re: 2015 permutation X2 marathon

new q:
There are 'n' cars in a parking lot with 'n' exits. Any car can go into any exit. What is the probability that there is at least one exit which is not accessed by any car?
Again, don't have the solution to this

The answer for anyone who wants to peek: 1 - n! / (n^n)

 

[kawaiipotato]

Re: 2015 permutation X2 marathon

The answer for anyone who wants to peek: 1 - n! / (n^n)

lol mine's way off. I did 1 - (n-1)^n / n^(n-1)

 

[braintic]

Re: 2015 permutation X2 marathon

lol mine's way off. I did 1 - (n-1)^n / n^(n-1)

See what happens with your answer when n=4.

 

[Drsoccerball]

Re: 2015 permutation X2 marathon

Probability of all cars going into one door = 1/n^n
Therefore 1- probability of each car going through a different gate: 1-(n!/n^n)

 

[braintic]

Re: 2015 permutation X2 marathon

Probability of all cars going into one door = 1/n^n
Therefore 1- probability of each car going through a different gate: 1-(n!/n^n)

The probability they all go through one door is actually 1 / [n^(n-1)]

Although you have my answer, I don't see how it follows from your first line.